Question

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot

is pulled away from the wall through a distance p so that its upper end slides a

distance q down the wall and then the ladder makes an angle β to the horizontal.

Show that

p/q = cosβ-cosα/sinα-sinβ​

Answer 1

Let QO = x and given that BQ = q, such that BO = q+x is the height of the wall.

Let AO = y and given that SA = q. ∠BAO = α and ∠QSO = β

In triangle BAO,

Sin a= BO/AB[ by property of sin theta=p/h]

=> BO= AB sin a...eq.1

Cos a = AO/AB[ by property of cos theta= b/h]

=>AO = AB Cos a...eq.2

In triangle QSO,

Sin B= QO/SQ[ by property sin theta = p/h]

=> QO= SQ Sin B...eq.3

Cos B = SO/SQ[ by property of cos theta = b/h]

=> SO= SQ Cos B...eq.4

Subtracting eq. 2 from eq. 4, we get SO –AO = SQ cos β – AB cos α ⇒ SA = SQ cos β – AB cos α

[from the above figure, SO –AO = SA = p] ⇒ p = AB cos β – AB cos α

[∵ SQ=AB=length of the ladder] ⇒ p = AB (cos β – cos α) …eq. 5

And subtracting eq. 3 from eq. 1, we get BO –QO = AB sin α – SQ sin β ⇒ BQ = AB sin α – SQ sin β

[from the above figure, BO –QO = BQ = q] ⇒ q = AB sin α – AB sin β

[∵ SQ=AB=length of the ladder] ⇒ q = AB (sin α – sin β) …eq. 6

Dividing eq. 5 and eq. 6, we get

=> p/q= AB(cos B- cos a )/ AB ( Sin a- Sin B)

=> p/q= Cos B- Cos a/ Sin a- Sin B

Hence Proved!!